find the distance of a point from a line?

**Miasmagasma** · December 11th 2009, 09:38 AM

i'm truly sorry if this is the wrong forum, as i'm teaching myself this stuff, i'm unsure of what it's sub forum should be. since my last question was moved here, i put this one here too.

anyway.....

use the following procedure to find the leats ( perpendicular ) distance of the point ( 1,2 ) from the line y = 3x + 5, without having to find the equation of a line perpendicular to y = 3x + 5.

(a) let ( x,y ) be a general point on the line, Show that its distance, d, from ( 1,2 ) is given by d^2 = ( x - 1 )^2 + ( y - 2 )^2.

(b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

(c) by completing the square, show that the minimum distance required 3/5√10

**Isomorphism** · December 11th 2009, 10:39 AM

Originally Posted by Miasmagasma

(b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x + 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

(c) by completing the square, show that the minimum distance required 3/5√10

$d^2 = ( x - 1 )^2 + ( 3x + 3 )^2$
$= 10x^2 + 16x + 10$
$= (\sqrt{10}x)^2 + 2\left(\frac{8}{\sqrt{10}}\right)(\sqrt{10}x) + \left(\frac{8}{\sqrt{10}}\right)^2 + \left(10 - \frac{64}{10}\right)$
$= \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 + \frac{36}{10}$
$\boxed{d^2 \geq \frac{36}{10}}$ since $\left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 \geq 0$

$d^2 \geq \frac{36}{10} \implies d \geq \frac{6}{\sqrt{10}}$

Now show that $\frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5}$ and you are done.

**Miasmagasma** · December 11th 2009, 10:52 AM

Originally Posted by Isomorphism

$d^2 = ( x - 1 )^2 + ( 3x + 3 )^2$
$= 10x^2 + 16x + 10$
$= (\sqrt{10}x)^2 + 2\left(\frac{8}{\sqrt{10}}\right)(\sqrt{10}x) + \left(\frac{8}{\sqrt{10}}\right)^2 + \left(10 - \frac{64}{10}\right)$
$= \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 + \frac{36}{10}$
$\boxed{d^2 \geq \frac{36}{10}}$ since $\left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 \geq 0$

$d^2 \geq \frac{36}{10} \implies d \geq \frac{6}{\sqrt{10}}$

Now show that $\frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5}$ and you are done.

can you explain how you changed it to this...

?

**Isomorphism** · December 11th 2009, 11:02 AM

Originally Posted by Miasmagasma

can you explain how you changed it to this...

?

That step is generally called completing the square. Read about it here.

**Miasmagasma** · December 11th 2009, 11:21 AM

Originally Posted by Isomorphism

That step is generally called completing the square. Read about it here.

**Miasmagasma** · December 11th 2009, 12:09 PM

i haven't been taught that method of completing the square in my book. is it anything to do with it being OCR board?

i have only learnt the ( x + 1/2b )^2 - 1/4b^2 + c

can that method be used as well?

**Miasmagasma** · December 11th 2009, 01:09 PM

could someone please walk me through this? i'm losing my marbles. i attempted the problem but i have no idea why he has square rooted all the 10s. if you are meant to divide through the whole equation why isn't it

x^2 + 16/10x + 1?????????????

**11rdc11** · December 11th 2009, 03:50 PM

Originally Posted by Miasmagasma

could someone please walk me through this? i'm losing my marbles. i attempted the problem but i have no idea why he has square rooted all the 10s. if you are meant to divide through the whole equation why isn't it

x^2 + 16/10x + 1?????????????

$\bigg(x+\frac{8}{5}\bigg)^2 +1 -\frac{64}{100}$

$\bigg(x+\frac{8}{5}\bigg)^2 +\frac{36}{100}$

Now remember you divided by 10 at the begining so you have to multiply 10 back in

$10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}$

so now you have

$d^2 = \bigg({10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}}\bigg)^2$

To solve for d take square root of both sides

$d = \sqrt{10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}}$

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